This booklet will discuss some of the principles involved in the design of a roller

coaster. It is intended for the middle or high school teacher. Physics students may find theinformation helpful as well. Many of the concepts can be applied to topics other than rollercoasters. Some sections will use the “Roller Coaster Simulator,” RCS. (See page 78 forinstructions on its construction.) The included activities are hands on cookbook type. Eachsection includes background topics that should have been taught previously.

This book assumes some rudimentary knowledge of physical science. It is a simplified

view of what design considerations and science a mechanical/civil design engineer must know when designing a roller coaster.

-1- by Tony Wayne

-2- by Tony Wayne A roller coaster is a balance between safety and sensation. Naturally, the ride shouldbe as safe as possible. After all, if the people are injured riding the coaster then therewould be fewer repeat riders. Fewer repeat riders means a short life span for the coaster.On the other hand, passengers ride a coaster for the “death defying” thrill. The key to asuccessful coaster is to give the rider the sensation of speed and acceleration. It all comesdown to speed control. To achieve this, the hills, curves, dips, straight always, braking systems and loops arenot randomly designed. They follow some simple rules of physics.

In order to understand what is going on, students must understand the differencebetween velocity and acceleration.

Velocity describes how quickly an object changes it position. The higher the velocitythe quicker an object travels between 2 locations. Phrases like, “...how fast..., how quickly,”are used to describe velocity. Often the word speed is substituted for the word velocity incommon usage. However, technically the two are different. Velocity is actually speed withdirection. For example, “60 mph, west,” is a velocity. “West” is the direction and “60 mph” isthe speed. The units of velocity are in the form of

Velocity = Units of Distance

Units of Time

Example = meter mile furlong

second hour week

Acceleration describes how quickly an object changes its velocity. Phrases like,“...slow down..., ...speed up..., ...change speed... and change velocity...” are used todescribe accelerations. If a student wants an easy way to determine if he is visualizingacceleration or a constant velocity along a straight line he only needs to ask himself onequestion, “Is the object slowing down or speeding up?” If the answer is “Yes,” then it isaccelerating. If the answer is “No” then it is moving with a constant velocity. The units ofacceleration are in the form of

Acceleration = Units of Distance = Units of Velocity

-3- by Tony Wayne

-4- by Tony Wayne The simple roller coaster started with Galileo Galilei. Below is a description of how todemonstrate Galileo’s experiment using the “Roller Coaster Simulator.” [This experimentcould also be duplicated using a HotWheels™ track and a marble.”

Marble STARTS here Adjust the ramp to lower angles. The marble keeps rolling to the same height. Galileo concluded from a similar experiment that the marble will keep rolling until it reaches the Marble STOPS here same height it started from if there were no friction. (With friction, it will keep rolling until it reaches the stop height.) Marble does not reach “stop height” and continues to move.

This has far-reaching implications. (1) The marble could take any path until it reachesthe same height it starts from, assuming no friction. In the previous activity, the marble did notroll to the same height it started from because of friction. But it consistently rolls to the sameheight. To reflect these implications, the track could be reshaped as shown below.

Marble STARTS here Marble STARTS here

Marble STOPS here Marble STOPS here

Marble STARTS here Marble does not make it Marble STARTS here Marble makes it around the around the loop segment. loop because it never reaches the stop height.

Marble STOPS here Marble STOPS here

-5- by Tony Wayne

(2) The ball begins to roll down due to the force of gravity. It stops when all the energy gravitygave the ball is used up. The marble accelerates only while a force acts on it in its direction ofmotion.

Here is a good exercise to draw on a chalk board.

The gray sections of track are where the coaster car accelerates. (Speeds up or slows down.)

The black sections of track are where the coaster car travels at a constant velocity.

The acceleration can be demonstrated experimentally using the roller coaster simulator orHotWheels™ track. If a long enough section is made horizontal, it can be shown that theaverage velocities calculated at the beginning and at the end of the horizontal section areequal. Form the track in the shape shown below. Roll a marble or steel bearing down thetrack. It will accelerate along the drop and move at a constant velocity along the horizontalsection and slow down as it climbs up the opposite side. When the marble slows down andspeeds up on the hills it is visually obvious. What is not so visually obvious is what happensalong the horizontal section of the track. The ball’s constant velocity can be shownmathematically. Divide the horizontal section of the track into 2 sections. Calculate theaverage velocity of the ball along these two sections. If done accurately, the velocities will benearly equal. To obtain more accurate results, use fairly long sections of horizontal track. Thelonger the sections of track, the greater the time measurement. Longer time measurementsmean lower percent errors.

-6- by Tony Wayne

Marble STARTS here Marble STOPS here

Force of gravity Force of gravity changes the changes the V1avg V2avg speed of the ball. speed of the ball. Velocity for the ball does not change

Along the horizontal section of the track, ignoring the minimal effects of friction, there are noforces acting on the ball horizontally. Therefore the ball moves at a constant velocity while noforce acts on it. This is Galileo’s law of inertia!!!

-7- by Tony Wayne

Here is another example of an illustration of Galileo’s experiment.

RING STAND WITH AN

ATTACHED PENDULUM.

RELEASE HEIGHT

AS THIS PEG IS MOVED

HIGHER AND LOWER, THE BALL WILL CONTINUE TO REACH THE HEIGHT IT IS DROPPED FROM.

RELEASE RELEASE RELEASE

HEIGHT HEIGHT HEIGHT

The ball continues until it reaches the starting height.

-8- by Tony Wayne

-9- by Tony Wayne A roller coaster is called a coaster because once it starts it coasts through the entiretrack. No outside forces are required for most coasters. (A few have double or triple lift hillsand braking sections.) Roller coasters trade height for velocity and velocity for height. Most allcalculations rely on using velocity measurements in one way or another. The first step isbeing able to calculate the changes in speed. In an ideal world, mechanical energy is conserved. Frictional forces are ignored inearly design stages. (This document does not address the nuances of dealing with frictionalforces.) Mechanical energy on a roller coaster comes in two basic forms. Kinetic energy, KE =(1/2)mv2 , and potential energy, PE = mgh, due to gravity. Total energy, ET, is conserved andis equal to the sum of kinetic and potential at any single location. ET = KE + PE (at any single location)

8.8 m/s 8.8 m/s

95 m 95 m 65 m 65 m

8.8 m/s 8.8 m/s

95 m 95 m 65 m 65 m

All four of these roller coasters look different. But because friction is ignored all four roller coasters will have the same speed at the bottom of each hill and at the top of the second. The only thing that matters is the height of the locations of interest.

- 10 - by Tony Wayne to calculate a change in velocity associated with a change in height Step 1 Identify two locations of interest. One with both a speed and a height and the other location with either speed or height. Step 2 Write an equation setting the total energy at one location equal to the total energy at the other location. Step 3 Solve for the unknown variable.

What is the velocity at the bottom of the first hill?

(1/2)v2 + gh = (1/2)mv2 + mgh The masses cancel out because it is the same coaster at the top and bottom. (1/2)v2 + gh = (1/2)v2 + gh Substitute the numbers at each location (1/2)(8.8)2 + 9.8(95) = (1/2)v2 + 9.8(0) The height at the bottom is zero because it is the lowest point when comparing to the starting height. 77.44 + 931 = ( 1/2)v2 1008.44 = ( 1/2)v2 2016.88 = v 2 ∴ v = 44.9 m/s ...at the bottom the the 1st hill.

What is the velocity at the top of the second hill?

ET(TOP OF 1st HILL) = ET(TOP OF 2nd HILL)

KE + PE = KE +PE (1/2)mv2 + mgh = (1/2)mv2 + mgh (1/2)v2 + gh = (1/2)mv2 + mgh The masses cancel out because it is the same coaster at the top and bottom. (1/2)v2 + gh = (1/2)v2 + gh Substitute the numbers at each location (1/2)(8.8)2 2 + 9.8(95) = ( /2)v + 9.8(65) 1 Notice all the numbers on the left side come from the top of the 1st hill while all the numbers on the right side come from the top of the 2nd hill. 77.44 + 931 = ( 1/2)v2 + 637

This technique can be used to calculate the velocity anywhere along the coaster.

- 11 - by Tony Wayne- 12 - by Tony Wayne Something has to be done to get the coaster started. In our previous example energy,power, has to added get the coaster up to 8.8 m/s. This is done by doing work on the coaster.A simplified definition of work would be force times displacement when the force anddisplacement go in the same direction. [This chapter will not go into all the details ofcalculating work.] Suffice it to say that when the force acting on the coaster and thedisplacement of the coaster are in the same direction, work adds energy to the coaster.When the force acting on the coaster and the displacement of the coaster are in oppositedirections, work removes energy from the coaster.

ET(BEGINNING) + Work = ET(TOP OF 1st HILL)

- 13 - by Tony Wayne 110,000 = ( 1/2)(3000)v2 73.333 = v 2 v = 8.6 m/s ... at the end of the catapult. As an aside you can calculate the acceleration of the rider from kinematics equations. (For the curious the acceleration is 7.0 m/s 2)

A roller coaster train of mass 3.0 X 103 kg rolls over a 11.5 m high hill at 8.34 m/s beforerolling down into the station. Once in the station, brakes are applied to the train to slow itdown to 1.00 m/s in 5.44 m.(a) What braking force slowed the train down?(b) How much time did it take to slow the train down?(c) What was the acceleration of the train in g’s?

Calculate the velocity as the train enters the station. Use this velocity to calculate the time.ET(HILL) = ET(@ STATION ENTRANCE) No work is done because no force acts between the twolocations KE + PE = KE +PE ( /2)mv2 + mgh = (1/2)mv2 + mgh 1

Calculate the acceleration in m/s 2. Then convert it into g’s.

- 15 - by Tony Wayne- 16 - by Tony WayneWeight is the pull of gravity. Typical weight units are pounds and newtons. (1 pound ≅4.45 newtons). On the moon, gravity pulls with 1/6 the force compared to the Earth. Therefore,a student on the moon weighs 1/6 of what she weighs on the Earth.On the Earth, neglecting air resistance, all objects will speed up at a rate of 9.80 m/s everysecond they fall. That is a speed increase of about 22 mph for every second an object falls.

Time in the Air Velocity

s mph

0 0 1 22 2 44 3 66 4 88 5 110

There are two ways to experience weightlessness. (1) move far enough away from theplanets and sun to where their pull is nearly zero. [Gravity acts over infinite distance. One cannever completely escape it.] (2) Fall down at a rate equal to the pull of gravity. In other words,accelerate to the Earth speeding up 22 mph every second in the air. In order for a person tofeel weight, a person must sense the reaction force of the ground pushing in the oppositedirection of gravity.

PULL OF GRAVITY (WEIGHT)

REACTION FORCE OF THE GROUND

In the absence of the reaction force a person will sink through the ground. Many amusement park rides generate the weightless sensation by accelerating downat 22 mph every second.

gÕs Neglecting air resistance, if a rock is dropped, it will accelerate down at 9.8 m/s 2 . Thismeans it will speed up by 9.8 m/s for every second it falls. If a rock you drop accelerates downat 9.8 m/s 2 , scientists say the rock is in a “1 g” environment, [1 g = 9.8 m/s 2 = 22 mph/s] . Any time an object experiences the pull equal to the force of gravity, it is said to be in a“one g” environment. We live in a 1 g environment. If a rock whose weight on the Earth is 100lbs was moved to a 2 g environment then it would weigh 200 lbs. In a 9 g environment itwould weigh 900 lbs. In a “NEGATIVE 2 g” environment it would take 200 lbs to hold the rockdown on the ground. In a “-5 g” environment it would take 500 lbs to hold the rock down to theground. If the rock were put into a “zero g” environment then it would be weightless. However,

- 17 - by Tony Wayneno matter what happens to its weight the rock’s mass would never change. Massmeasurement is unaffected by the pull of gravity. What does it feel like to walk in a 2 g environment? Have students find someone who’smass is about equal to theirs. Have them give piggyback rides. As they walk around this iswhat it feels like to be in a 2 g environment. Go outside on the soft ground and have thestudents step up on something. This is when they will really know what a 2 g environmentfeels like. Often engineers will use g’s as a “force factor” unit. The force factor gives a person away of comparing what forces feel like. All acceleration can be converted to g’s by dividing the answer, in m/s 2, by 9.8 m/s 2.

A roller coaster is propelled horizontally by a collection of linear accelerator motors.

The mass of the coaster train is 8152 kg. The train starts from rest and reaches a velocity of26.1 m/s , 55 mph, in 3.00 seconds. The train experiences a constant acceleration. What is thecoaster train’s acceleration in g’s?

The rider is pushed back into the

BACKGROUND Cut the rubber bands. Tie the ends of each together to make a stretchy string. Tie the weights to the opposite ends of the rubber band. Attach the middle of the rubber band to the inside bottom of the cup. The two masses should be able to hang over the lip of the cup.

RUBBER BAND

Tension in the rubber band Normal force 50 g mass exerted by the cup on the weight weight (mg) Free body diagram of the forces acting on CUP the left hanging weight.

The masses are in equilibrium with the upward force of the rubber band. The force pulling up of the rubber bands is equal to the force of gravity, [weight of the masses.] Ask the class, “What would happen if the rubber bands pulled with a force greater than the pull of gravity on the masses?” The masses would shoot upward and be pulled into the cup. To show this, pull down on one of the masses and let go.

Now ask, “What would happen if the masses could be magically shielded from the pull of gravity?” With no force stretching the rubber bands, they would sling shot into the cup.

Explain, “We cannot yet shield gravity. But we can momentarily minimize its effects by accelerating the masses, rubber band and cup down at 9.80 m/s 2, the acceleration due to gravity. Without saying anymore, stand on a chair. Raise the apparatus with the weights hanging out. Tell the students, “When this cup is dropped everything will speed up equal to the acceleration of gravity. What will you see when this cup is dropped?” Drop the cup after polling the students. The masses will be pulled into the cup. When everything falls, gravity will not be pulling against the masses when compared to the - 19 - by Tony Wayne rubber band’s pulling force. The masses are said to be weightless. It is the weight of the mass that stretched the rubber band. If the mass is weightless, the rubber band will pull it in.

Drop

Continue to the ground

When everything is falling together, the pull of gravity is no longer experienced by therubber bands. Therefore, they pull the masses into the cup.

STEP 1 STEP 2 STEP 3 STEP 4

Nail it together. Glue and nail the Attach the nail to Insert the balloon triangles to the the weight and in the hole. Inflate outside. hang the weight it so that it does by rubber bands. not touch the nail. Tie it closed

The DEMO With the balloon inflated, hold the frame over a pillow. Hold the frame straight out at chest height. Ask students to predict what will happen when you release the frame. Guide them to specifics such as where in the fall will the balloon pop. Release the frame. The balloon will pop almost instantaneously. The balloon pops because the weighted pin becomes weightless. The rubber bands are essentially pulling against nothing. This means the rubber bands pull the pin up into the balloon.

Pull of the rubber Pull of the rubber

bands bands

When falling, the rubber

Pull of gravity bands are moving down due to the force of gravity. this motion negates the pull of gravity on the hanging mass. The rubber Forces acting on the mass when it is at rest. bands can now pull the hanging mass up -because it is weightless in free fall

PREPARATION Heat up the nail with the candle flame. Be careful not to burn yourself. Poke the hot nail into opposite sides of the cup at the bottom. This will make a clean hole. Hold your fingers over the two holes and fill the cup half full of water.

THE DEMO Stand on a chair and briefly release your fingers from the holes. The water should stream out the cup’s holes onto the floor. Ask the students, “What will happen when I drop the cup into the trash can?” Listen to all their answers. Drop the cup to see who’s prediction was correct. The water will not flow out of the cup. Water flows out of the cup when the acceleration of the cup is different from the water. When the cup is held the water is allowed to accelerate down at 9.8 m/s 2. When the cup falls too, the cup is also accelerating down at 9.8 m/s 2. Since there is no difference in their accelerations the water stays in the cup.

Force holding Remove the force holding the cup up

the cup up. by letting go of the cup.

CUP WITH CUP WITH

2 HOLES 2 HOLES .

Pull of gravity Pull of gravity

The water is accelerating down

The water is not falling down quicker faster than the the cup. (That is than the cup, so it stays in the cup. because the cup is not accelerating at all.)

- 22 - by Tony Wayne- 23 - by Tony WayneOne of the most basic parts of a ride is going from the top of a hill to the bottom. There aretwo basics ways designs transport riders to the bottom of a hill. The first is called the“Speed Run.”

vo vo = velocity at the hill’s top vf = velocity at the hill’s bottom h = the height of the drop g = the acceleration due to gravity (9.80 m/s 2)

vf = vo2 +2gh

vf

A speed run is designed to give the rider the feeling of accelerating faster and fasterwithout the feeling of weightlessness. It simulates being in a powerful car with theaccelerator held down to the floor. It is a straight piece of track that connects a high pointto a low point. The increase in velocity of the car comes from lost gravitational potential energy beingconverted into kinetic energy. Next to a horizontal straight piece, the speed run is theeasiest piece of the track to design and analyze.

When coasting up to a new height the calculations are the same as the example shownabove. The shape of the hill does not matter. See the “Intro to Design” section, step 7, foran example of these up hill calculations.

- 24 - by Tony Wayne One of the biggest thrills on a roller coaster is the free fall as a rider travels over a hill.The easiest way to experience free fall is to hang from a tall height and drop to theground. As a person falls he experiences weightlessness. As long as a person travels inthe air like a projectile he will feel weightless. Suppose a ball traveled off a table, horizontally, at 10 m/s . The ball’s path would looklike the path shown below.

BALL IN FREE FALL

Now suppose the ball traveled off the table top on a shallow angled ramp. It wouldlook like the one below.

The ball will take longer to descend. The ramp

applies an upward force on the ball to change its path and thereby slow its descent. To the ball, this upward force is registered as weight.

Where the ball

would be without the ramp.

For this section of the trip, the ball is weightless.

For this section of the

trip, the ball feels weight.

Where the ball

would be without the ramp.

- 25 - by Tony Wayne A straight, “speed run,” drop does not match the fall of a rider over a hill.

For this section of the trip, the ball is weightless.

Not the safest of choices if the ball were a roller coaster car full of passengers.

To give riding more of a thrill, the designer needs to design the shape of the hill to match thefalling ball.

BALL IN FREE FALL

The hill is the same shape as a projectile in free fall. The rollercoaster barely makes contact with the track.

The only problem with curve above is the impact with the floor. To alleviate this problem another curve scoops the balls as they descend. This makes the ride smooth and survivable for the rider.

BALL IN FREE FALL

The speed at the bottom of a free fall drop is calculated the same way as the speed at thebottom of a speed run drop. The only difference is the shape of the hill from the top to thebottom.

- 26 - by Tony Wayne A free fall hill shape gives a rider a weightless sensation. To give this weightless sensation over a hill, the hill is designed to have the same shape as the path of a ball being thrown off the top of a hill. Shape is determined by how fast the roller coaster car travels over the hill. The faster the coaster travels over the hill the wider the hill must be. There are two ways to apply projectile motion concepts to design the hill’s shape. The first way is to calculate the coaster’s position as if it drove off a cliff. The position equation is as follows.

h = the height from the top of the hill

x = is the distance away from the center of the hill v = the velocity the roller coaster car travels over the top of the hill. g = the acceleration due to gravity. 9.80 m/s 2 for answers in meters. 32.15 ft /s 2 for answers in feet.This can be rewritten as

There comes a certain point on the free-fall drop where the track needs to redirect the riders. Otherwise the riders will just plummet into the ground. This point is the transition point from free-fall to controlled acceleration. This point is also the maximum angle of a hill. This angle can be in virtually any range from 35° to 55°. - 27 - by Tony Wayne Tangent line whose slope is 50°

Transition point where the slope of

the free-fall hill equals 50°

This section of the track will not

follow the equation for free fall.

For the bottom section of the track, the new equation has the desired outcome of changing the direction of the coaster from a downward motion to a purely horizontal motion. The track will need to apply a vertical component of velocity to reduce the coaster’s vertical velocity to zero. The track will also need to increase the horizontal velocity of the coaster to the value determined from energy relationships. The velocity at the bottom of the hill is determined from

Total mechanical Total mechanical energy at the top energy at the bottomwhich is (1/2)m(vT)2 + (mgh) = (1/2)m(vB)2This simplifies to vB = (vT)2 + 2ghwhere vB is the horizontal velocity at the bottom of the hill. The value for vB will be used in later calculations.

Where “vy ” is the final vertical velocity of zero, “vyo ” is the vertical component of the velocity at the transition point, and “y” is the distance left to fall from the transition point to the ground. The horizontal velocity is determined from a parameter decided upon by the engineer. The engineer will want to limit the g forces experienced by the rider. This value will be the net g’s felt by the rider. These net g’s are the net acceleration.

anet ay

ax ∴ anet 2 = ax 2 + ay 2

and

ax = anet 2 - ay 2these values are plugged back into the original equation and x values are calculated as a function of y.

- 29 - by Tony WayneThis is the beginning of the Hurler at Paramount’sKings Dominion in Doswell Virginia. Can you tellwhich hill is the free fall hill?

(It’s the curved hill in front.)

- 30 - by Tony WayneA person throws 2 balls. The first ball is thrown horizontally at 20 m/s. The second ball is thrown at 40 m/s. Draw as much ofeach path as possible. Draw the ball’s position every 20 m of vertical flight. Draw a smooth line to show the curve’s shape.

0 20 40 60 80 100

200

300

400

Vertical Horizontal Horizontal

distance down position for position for (meters) the ball with the ball with an initial an initial velocity of 20 velocity of 40 m/s m/s

0 0 0 40 80 120 160 200 240 280 320 360 400

- 31 - by Tony WayneA person throws 2 balls. The first ball is thrown horizontally at 10 m/s. The second ball is thrown at 30 m/s. Draw as much ofeach path as possible. Draw the ball’s position every 1 second of the flight. Draw a smooth line to show the curve’s shape.

0 20 40 60 80 100

200

300

400

Time Vertical Vertical

(seconds) position for position for the ball with the ball with an initial an initial velocity of 10 velocity of 30 m/s m/s

0 0 0 1 2 3 4 5 6 7 8 9

- 32 - by Tony WayneA person throws 2 balls. The first ball is thrown horizontally at 20 m/s. The second ball is thrown at 40 m/s. Draw as much ofeach path as possible. Draw the ball’s position every 20 m of vertical flight. Draw a smooth line to show the curve’s shape.

0 20 Velocity = 40 m/s 40 60 80 100

200

Velocity = 20 m/s

300

400 0 100 200 300 400 500 600 Horizontal Distance in meters

Below is a set of answers for 4 velocities.

Vertical Horizontal Horizontal Horizontal Horizontal distance down position for position for position for position for (meters) the ball with the ball with the ball with the ball with an initial an initial an initial an initial velocity of 10 velocity of 20 velocity of 30 velocity of 40 m/s m/s m/s m/s

To see the effect of speed on arc shape.

Hold the end of the ruler up to three different angles. Repeatedly roll the ball down the ruler until you can draw the path the ball takes for each height. (Make sure the marble does not bounce when it leaves the table.) Write your data down for each trial in the data table below.

V L H D Length of the ramp the Height where the ball Distance to the ball’s Average velocity along ball rolls down. starts landing spot. the table top ONLY. (m) (m) (m) (m/s)

Vavg = dist /time

Draw the path the ball takes on the grid on the next page for each ruler height.

- 34 - by Tony Wayne RULER Make sure the ball does not bounce when it leaves the 1m table.

The drawn path represents the shape of a free fall roller coaster hill for different coaster carvelocities.

1 Which ramp angle gives the ball the greatest speed when it leaves the table?

2 What conclusions can you make about how speed over the top of a hill affects the shape of the hill?

- 35 - by Tony Wayne Below are three different starting heights for a hill. For each starting height, rate the hills shape as:(1) Possible optimum hill. The ball will be in free fall the longest period of time and the ball would softly be caught at the bottom.(2) Safe hill. The ball stays on the hill but does not remain in free fall for the maximum amount of time.(3) Unsafe hill. The ball will leave the track and possibly hit the other side of the track.

STARTING HEIGHT #1

STARTING HEIGHT #2

STARTING HEIGHT #3

3 - 36 - by Tony WayneIn this activity the student will visualize the path of a ball as it rolls over different shaped hills.In this activity a ball will be rolled from 3 different heights over three different shaped hills.• Lay out the roller coaster simulator track on the chalk board.• Using a piece of chalk trace the path of the track and mark the starting point of the track.• Roll the ball down the track. If the ball leaves the track, trace its path with chalk. Then roll it again to see how it compares with the drawn line.• Draw the path the ball takes on the paper. Repeat this process for every hill and every starting position. Starting height 1 HILL SHAPE “A”

Starting height 1 HILL SHAPE “B”

Starting height 1 HILL SHAPE “C”

- 37 - by Tony Wayne HILL SHAPE “A”Starting height 2

HILL SHAPE “B”

Starting height 2

HILL SHAPE “C”

Starting height 2

- 38 - by Tony Wayne HILL SHAPE “A”

Starting height 3

HILL SHAPE “B”

Starting height 3

HILL SHAPE “C”

Starting height 3

- 39 - by Tony WayneBelow are three different starting heights for a hill. For each starting position rate the hillsshape as:(1) Possible optimum hill. The ball will be in free fall the longest period of time and the ball would softly be caught at the bottom.(2) Safe hill. The ball stays on the hill but does not remain in free fall for the maximum amount of time.

(3) Unsafe hill. The ball will leave the track and possibly hit the other side of the track.

1 2 3 - 40 - by Tony WayneStarting height 1 Hill shape “A” Starting height 2 Starting height 3 All these starting heights will keep the marble on the track. (Starting height 3 may not even make it over the track.)

Starting heights 2 & 3

- 41 - by Tony Wayne The ball is the highest The ball will get airborne The ball will get airborne

1 and will be traveling the

fastest. A fast speed means a long “flat” hill. over this hill. over this hill.

The ball will make it over The ball is the highest

2 the hill. The ball will not and will be traveling the experience weightlessness fastest. A fast speed means because the ball wants to fall a long “flat” hill. The ball will get airborne over this hill.

quicker than the hill will allow.

The ball will make it over The ball will make it over The ball is the highest

3 the hill. The ball will not

experience weightlessness because the ball wants to fall the hill. The ball will not and will be traveling the experience weightlessness fastest. A fast speed means because the ball wants to fall a long “flat” hill. quicker than the hill will allow. quicker than the hill will allow.